I would like to verify the proof given in an answer #2 form here. The claim to prove was that $\lambda$ is an eigenvalue of $T$ if and only if $\overline{\lambda}$ is eigenvalue of the adjoint operator $T^*$ (I assume finite dimensional inner product space). Since:
$\langle T^*u, v \rangle = \overline{\langle Tv, u \rangle } = \overline{\langle\lambda v, u \rangle } = \overline{\overline{\lambda}\langle v,u\rangle } = \lambda\langle u,v\rangle = \langle\overline{\lambda} u,v \rangle \space\quad (1)$
and similarly
$\langle Tv, u \rangle = \overline{\langle T^*u,v\rangle} = \overline{\langle \overline{\lambda}u,v \rangle} = \overline{\lambda \langle u,v \rangle} = \overline{\lambda}\langle v,u\rangle = \langle \lambda v, u\rangle \quad(2)$
The answer seems (to me) to imply that since $\langle T^*u, v \rangle = \langle\overline{\lambda}u,v \rangle$, this implies $ T^*u = \overline{\lambda}u \ $ from $(1)$. But isn't it the case that this is true only if it holds for all vectors $v$, whereas $v$ in $(1)$ is an eigenvector? And similarly isn't it the case that $\langle Tv, u \rangle = \langle \lambda v,u \rangle$ implies $Tv=\lambda v$ only if it holds for all vectors $u$?
